Mysterious Planet: Approximating its Density

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SUMMARY

The discussion centers on calculating the density of a newly discovered planet with no atmosphere, based solely on the orbital period of its satellite, which is 1.50 hours. Participants emphasize the importance of equating gravitational force to centripetal force using the formula mGM/R^2 = mω^2R. They derive the density formula as ρ = (1/G)(2π/T)^2, indicating that the density can be determined from the orbital period without needing the planet's radius or mass directly.

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  • Understanding of gravitational force and centripetal force concepts
  • Familiarity with the formula for density ρ = M/V
  • Knowledge of angular velocity and its relationship to orbital period
  • Basic grasp of mathematical manipulation of physical equations
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  • Study the derivation of gravitational force and centripetal force equations
  • Learn about the implications of Kepler's laws on orbital mechanics
  • Explore the concept of density in astrophysics and its calculation methods
  • Investigate the role of gravitational constant G in celestial mechanics
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Astronomers, astrophysicists, and students of physics interested in celestial mechanics and density calculations of planetary bodies.

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A strange new planet that has no atmosphere has a satellite that orbits very close to the planet's surface with a period of 1.50 hours. What is the approximate density of the planet? (Assume that the radius of orbit equals the radius of the planet.)
 
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Maybe I'm wrong...but it sounds like something is missing here:
It only gives you the time it takes to complete a revolution?
I would think u either need to know the planet's radius, mass, or ship velocity to calculate the density...

Because, say a person drives a car around the Earth's equator, and someone else flies a plane around the same distance. Without knowing velocity of the objects, the period of each revolution would be different, even though its the same planet, hence same density...
 
How about just equating the gravitational force to the centripetal force?
[itex]\frac{mGM}{R^2} = m\omega^2R[/itex]

and since [itex]\rho = \frac{M}{R^3}[/itex] and v = Rω and [itex]T = \frac{2\pi}{\omega}[/itex] you can solve for density as a function of period. So I think:

[itex]\rho = \frac{1}{G}\left(\frac{2\pi}{T}\right)^2[/itex]
 

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